Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 - 6z - 40}{z + 4} \times \dfrac{z + 1}{z - 10} $
Answer: First factor the quadratic. $x = \dfrac{(z - 10)(z + 4)}{z + 4} \times \dfrac{z + 1}{z - 10} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (z - 10)(z + 4) \times (z + 1) } { (z + 4) \times (z - 10) } $ $x = \dfrac{ (z - 10)(z + 4)(z + 1)}{ (z + 4)(z - 10)} $ Notice that $(z + 4)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ \cancel{(z - 10)}(z + 4)(z + 1)}{ (z + 4)\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $x = \dfrac{ \cancel{(z - 10)}\cancel{(z + 4)}(z + 1)}{ \cancel{(z + 4)}\cancel{(z - 10)}} $ We are dividing by $z + 4$ , so $z + 4 \neq 0$ Therefore, $z \neq -4$ $x = z + 1 ; \space z \neq 10 ; \space z \neq -4 $